Bounding the Inefficiency of the Reliability-Based Continuous Network Design Problem Under Cost Recovery

Abstract

This study defines the price of anarchy for general reliability-based transport network design problems, which is an indicator of inefficiency that reveals how much the design objective value exceeds its theoretical minimum value due to the risk averse and selfish routing behavior of travelers. This study examines a new problem, which is a reliability-based continuous network design problem under cost recovery. In this problem, the variations of system travel time and path travel times, the risk attitudes of the system manager and travelers, congestion toll charges, capacity expansions, and cost recovery constraint are explicitly considered. The design problem is formulated as a min-max problem with the reliability-based user equilibrium constraint. It is proved that the price of anarchy for this problem is bounded above, and the upper bound is independent of travel time functions, demands, and network topology. The upper bound is related to the travel time variations, the value of reliability, and the value of time.

Appendices

Appendix 1: Lemma 1 and its Proof

Lemma 1

For any link flow pattern ${\mathbf{v}}^{\prime }={\left({v}_a^{\prime}\right)}_{a\in A}$, the following inequality holds:

$$ {\displaystyle \begin{array}{l}\sum \limits_{a\in A}\left({R}^{\mathrm{t}}{t}_a\left({v}_a^{RUE},{y}_a^{RUE}\right)+{\tau}_a\Big({v}_a^{RUE},{y}_a^{RUE}\left)-{R}^{\mathrm{t}}{t}_a\right({v}_a^{\prime },{y}_a^{RUE}\Big)\right){v}_a^{\prime}\\ {}\le \sum \limits_{a\in A}{\tau}_a\left({v}_a^{RUE},{y}_a^{RUE}\right)\cdot {v}_a^{RUE},\end{array}} $$

(20)

where ${v}_a^{RUE}$ and ${y}_a^{RUE}$ denote the entries of v^RUE(f^RUE) and y^RUE, respectively.

Proof

For an individual link a ∈ A, consider the following maximization problem:

$$ \underset{x_a\ge 0}{\min }{\overline{Z}}_a\left({x}_a\right)={R}^{\mathrm{t}}\left({t}_a\left({v}_a^{RUE},{y}_a^{RUE}\right)+{v}_a^{RUE}{dt}_a\Big({v}_a^{RUE},{y}_a^{RUE}\left)/{dv}_a-{t}_a\right({x}_a,{y}_a^{RUE}\Big)\right){x}_a. $$

The first order derivative of ${\overline{Z}}_a\left({x}_a\right)$ with respect to x_a is

$$ {\displaystyle \begin{array}{l}d{\overline{Z}}_a\left({x}_a\right)/{dx}_a=\\ {}{R}^{\mathrm{t}}\left({t}_a\left({v}_a^{RUE},{y}_a^{RUE}\right)+{v}_a^{RUE}{dt}_a\Big({v}_a^{RUE},{y}_a^{RUE}\left)/{dv}_a-{t}_a\right({x}_a,{y}_a^{RUE}\left)-{x}_a{dt}_a\right({x}_a,{y}_a^{RUE}\Big)/{dv}_a\right).\end{array}} $$

Because of the properties of the link travel time function and the marginal link cost toll function, the following hold: $d{\overline{Z}}_a\left({x}_a\right)/{dx}_a>0$ for $0\le {x}_a<{v}_a^{RUE}$; $d{\overline{Z}}_a\left({x}_a\right)/{dx}_a=0$ for ${x}_a={v}_a^{RUE}$, and $d{\overline{Z}}_a\left({x}_a\right)/{dx}_a<0$ for ${x}_a>{v}_a^{RUE}$. The objective function ${\overline{Z}}_a\left({x}_a\right)$ is strictly increasing on $\left[0,{v}_a^{RUE}\right]$ and strictly decreasing on $\left({v}_a^{RUE},+\infty \right)$. If ${v}_a^{RUE}$ equals zero, $d{\overline{Z}}_a\left({x}_a\right)/{dx}_a=0$ at x_a = 0 and $d{\overline{Z}}_a\left({x}_a\right)/{dx}_a<0$ for x_a > 0. The function ${\overline{Z}}_a\left({x}_a\right)$ is strictly decreasing on [0, +∞). The global maximum point ${x}_a^{\ast }$ of the objective function exists and is unique, and satisfies the condition: $d{\overline{Z}}_a\left({x}_a^{\ast}\right)/{dx}_a=0$, i.e., ${x}_a^{\ast }>{v}_a^{RUE}$.

Substituting the global maximum point ${x}_a^{\ast }$ into the objective function ${\overline{Z}}_a\left({x}_a\right)$, the maxima of the objective function is

$$ {\displaystyle \begin{array}{c}{\overline{Z}}_a\left({x}_a^{\ast}\right)={R}^{\mathrm{t}}\left({t}_a\left({v}_a^{RUE},{y}_a^{RUE}\right)+{v}_a^{RUE}{dt}_a\Big({v}_a^{RUE},{y}_a^{RUE}\left)/{dv}_a-{t}_a\right({v}_a^{RUE},{y}_a^{RUE}\Big)\right){v}_a^{RUE}\\ {}={R}^{\mathrm{t}}{v}_a^{RUE}{v}_a^{RUE}\cdot {dt}_a\left({v}_a^{RUE},{y}_a^{RUE}\right)/{dv}_a={\tau}_a\left({v}_a^{RUE},{y}_a^{RUE}\right)\cdot {v}_a^{RUE}.\end{array}} $$

Thus, given a feasible link flow ${v}_a^{\prime }$, the following inequality holds:

$$ {\displaystyle \begin{array}{c}{\overline{Z}}_a\left({v}_a^{\prime}\right)={R}^{\mathrm{t}}\left({t}_a\left({v}_a^{RUE},{y}_a^{RUE}\right)+{v}_a^{RUE}{dt}_a\Big({v}_a^{RUE},{y}_a^{RUE}\left)/{dv}_a-{t}_a\right({v}_a^{\prime },{y}_a^{RUE}\Big)\right){v}_a^{\prime}\\ {}=\left({R}^{\mathrm{t}}{t}_a\left({v}_a^{RUE},{y}_a^{RUE}\right)+{\tau}_a\Big({v}_a^{RUE},{y}_a^{RUE}\left)-{R}^{\mathrm{t}}{t}_a\right({v}_a^{\prime },{y}_a^{RUE}\Big)\right){v}_a^{\prime}\\ {}\le {\tau}_a\left({v}_a^{RUE},{y}_a^{RUE}\right)\cdot {v}_a^{RUE}.\end{array}} $$

(21)

Condition (21) holds for any individual link in the road network. Summing up condition (21) over all links on a path, the result (20) in the lemma is obtained.

Appendix 2: Upper Bounds of TSTCB and Sum of Individual Path Travel Cost Budgets

Based on the formula relating the path and link travel time standard deviation, the path travel time standard deviation is smaller than or equal to the sum of link travel time standard deviations of links on that path, i.e., ${\varsigma}_p\le \sum \limits_{a\in A}{\sigma}_a{\delta}_p^a$. Similarly, $\sigma \left[ TSTT\right]\le \sum \limits_{a\in A}{\sigma}_a{v}_a$. According to the definition of ε_max, we have ${\varsigma}_p\le \sum \limits_{a\in A}{\varepsilon}_{\mathrm{max}}{t}_a{\delta}_p^a$ and $\sigma \left[ TSTT\right]\le \sum \limits_{a\in A}{\varepsilon}_{\mathrm{max}}{t}_a{v}_a$.

Because $\sigma \left[ TSTT\right]\le \sum \limits_{a\in A}{\varepsilon}_{\mathrm{max}}{t}_a{v}_a$, it can easily be proved that the TSTCB has an upper bound, which is the mean TSTT multiplied by a number:

$$ {TSTCB}_{R^{\mathrm{t}},{R}^{\mathrm{s}}}\left(\mathbf{v}\left(\mathbf{f}\right),\mathbf{y}\right)\le \left({R}^{\mathrm{t}}+{\varepsilon}_{\mathrm{max}}{R}^{\mathrm{s}}\right)\sum \limits_{a\in A}{t}_a\left({v}_a,{y}_a\right)\cdot {v}_a. $$

(22)

Similar to the sum of individual path travel cost budgets, we have

$$ \sum \limits_{p\in P}{f}_p{b}_p\left(\mathbf{v}\left(\mathbf{f}\right),\mathbf{y}\right)\le \left({R}^{\mathrm{t}}+{\varepsilon}_{\mathrm{max}}{R}^{\mathrm{u}}\right)\sum \limits_{a\in A}{t}_a\left({v}_a,{y}_a\right)\cdot {v}_a. $$

(23)

Note that $\sum \limits_{p\in P}{f}_p{b}_p\left(\mathbf{v}\left(\mathbf{f}\right),\mathbf{y}\right)={R}^{\mathrm{t}}\sum \limits_{a\in A}{t}_a\left({v}_a,{y}_a\right)\cdot {v}_a+{R}^{\mathrm{u}}\sum \limits_{p\in P}{f}_p\cdot {\varsigma}_p$.

Appendix 3: Proof of Property 1

Proof

Assume ${\mathbf{v}}^{{\prime\prime\prime}}\left({\mathbf{f}}^{{\prime\prime\prime}}\right)={\left({v}_a^{{\prime\prime\prime}}\right)}_{a\in A}$ is the link flow pattern of the path flow pattern ${\mathbf{f}}^{{\prime\prime\prime} }={\left({f}_p^{{\prime\prime\prime}}\right)}_{p\in P}$ that minimizes the sum of individual path travel cost budgets. Let ${\varsigma}_p^{RUE}$ and ${\varsigma}_p^{{\prime\prime\prime} }$ be the path travel time standard deviations of f^RUE and f^‴, respectively. Let ${\tau}_a^{RUE}={\tau}_a\left({v}_a^{RUE},{y}_a^{RUE}\right)$, ${\tau}_a^{{\prime\prime\prime} }={\tau}_a\left({v}_a^{{\prime\prime\prime} },{y}_a^{RUE}\right)$, ${t}_a^{RUE}={t}_a\left({v}_a^{RUE},{y}_a^{RUE}\right)$, ${t}_a^{{\prime\prime\prime} }={t}_a\left({v}_a^{{\prime\prime\prime} },{y}_a^{RUE}\right)$, ${q}_p^{RUE}=\sum \limits_{a\in A}{t}_a^{RUE}{\delta}_p^a$, ${q}_p^{{\prime\prime\prime} }=\sum \limits_{a\in A}{t}_a^{{\prime\prime\prime} }{\delta}_p^a$, ${\tilde{b}}_p^{RUE}={b}_p^{RUE}+\sum \limits_{a\in A}{\tau}_a^{RUE}{\delta}_p^a$, and ${{\tilde{b}}^{{\prime\prime\prime}}}_p={b}_p^{{\prime\prime\prime} }+\sum \limits_{a\in A}{\tau}_a^{{\prime\prime\prime} }{\delta}_p^a$.

Because f^RUE is the RUE path flow pattern, the following inequality holds: $\sum \limits_{p\in P}\left({f}_p^{{\prime\prime\prime} }-{f}_p^{RUE}\right){\tilde{b}}_p^{RUE}\ge 0$ (for details, see the solution method in Sub-section 2.7 in the study of Szeto and Wang 2016), which is equivalent to $\sum \limits_{p\in P}{\tilde{b}}_p^{RUE}{f}_p^{RUE}\le \sum \limits_{p\in P}{\tilde{b}}_p^{RUE}{f}_p^{{\prime\prime\prime} }$. Subtracting $\sum \limits_{p\in P}{{\tilde{b}}^{{\prime\prime\prime}}}_p{f}_p^{{\prime\prime\prime} }$ from both sides of the above inequality, we obtain $\sum \limits_{p\in P}{\tilde{b}}_p^{RUE}{f}_p^{RUE}-\sum \limits_{p\in P}{{\tilde{b}}^{{\prime\prime\prime}}}_p{f}_p^{{\prime\prime\prime}}\le \sum \limits_{p\in P}\left({\tilde{b}}_p^{RUE}-{{\tilde{b}}^{{\prime\prime\prime}}}_p\right){f}_p^{{\prime\prime\prime} }$, which can be rewritten as

$$ {\displaystyle \begin{array}{l}\sum \limits_{p\in P}{b}_p^{RUE}{f}_p^{RUE}+\sum \limits_{p\in P}\left({\sum}_{a\in A}{\tau}_a^{RUE}{\delta}_p^a\right){f}_p^{RUE}-\sum \limits_{p\in P}{b}_p^{{\prime\prime\prime} }{f}_p^{{\prime\prime\prime} }-\sum \limits_{p\in P}\left({\sum}_{a\in A}{\tau}_a^{RUE}{\delta}_p^a\right){f}_p^{{\prime\prime\prime}}\le \\ {}{R}^{\mathrm{t}}\sum \limits_{p\in P}\left({q}_p^{RUE}-{q}_p^{{\prime\prime\prime}}\right){f}_p^{{\prime\prime\prime} }+{R}^{\mathrm{u}}\sum \limits_{p\in P}\left({\varsigma}_p^{RUE}-{\varsigma}_p^{{\prime\prime\prime}}\right){f}_p^{{\prime\prime\prime} }+\sum \limits_{p\in P}\left({\sum}_{a\in A}{\tau}_a^{RUE}{\delta}_p^a-{\sum}_{a\in A}{\tau}_a^{{\prime\prime\prime} }{\delta}_p^a\right){f}_p^{{\prime\prime\prime} },\end{array}} $$

or

$$ {\displaystyle \begin{array}{c}\sum \limits_{p\in P}{b}_p^{RUE}{f}_p^{RUE}-\sum \limits_{p\in P}{b}_p^{{\prime\prime\prime} }{f}_p^{{\prime\prime\prime}}\\ {}\le \left[{R}^{\mathrm{t}}\sum \limits_{p\in P}\left({q}_p^{RUE}-{q}_p^{{\prime\prime\prime}}\right){f}_p^{{\prime\prime\prime} }+\sum \limits_{p\in P}\left({\sum}_{a\in A}{\tau}_a^{RUE}{\delta}_p^a\right){f}_p^{{\prime\prime\prime} }-\sum \limits_{p\in P}\left({\sum}_{a\in A}{\tau}_a^{RUE}{\delta}_p^a\right){f}_p^{RUE}\right]\\ {}+{R}^{\mathrm{u}}\sum \limits_{p\in P}\left({\varsigma}_p^{RUE}-{\varsigma}_p^{{\prime\prime\prime}}\right){f}_p^{{\prime\prime\prime} }.\end{array}} $$

(24)

For the term in the square bracket on the right side of (24), we have: ${R}^{\mathrm{t}}\sum \limits_{p\in P}\left({q}_p^{RUE}-{q}_p^{{\prime\prime\prime}}\right){f}_p^{{\prime\prime\prime} }={R}^{\mathrm{t}}\sum \limits_{a\in A}\left({t}_a^{RUE}-{t}_a^{{\prime\prime\prime}}\right){v}_a^{{\prime\prime\prime} }$, $\sum \limits_{p\in P}\left({\sum}_{a\in A}{\tau}_a^{RUE}{\delta}_p^a\right){f}_p^{{\prime\prime\prime} }=\sum \limits_{a\in A}{\tau}_a^{RUE}{v}_a^{{\prime\prime\prime} }$, and $\sum \limits_{p\in P}\left({\sum}_{a\in A}{\tau}_a^{RUE}{\delta}_p^a\right){f}_p^{RUE}=\sum \limits_{a\in A}{\tau}_a^{RUE}{v}_a^{RUE}$. Thus, the term in the square bracket in (24) can be expressed in terms of link-based variables:

$$ {\displaystyle \begin{array}{l}\left[{R}^{\mathrm{t}}\sum \limits_{p\in P}\left({q}_p^{RUE}-{q}_p^{{\prime\prime\prime}}\right){f}_p^{{\prime\prime\prime} }+\sum \limits_{p\in P}\left({\sum}_{a\in A}{\tau}_a^{RUE}{\delta}_p^a\right){f}_p^{{\prime\prime\prime} }-\sum \limits_{p\in P}\left({\sum}_{a\in A}{\tau}_a^{RUE}{\delta}_p^a\right){f}_p^{RUE}\right]=\\ {}\sum \limits_{a\in A}\left({R}^{\mathrm{t}}{t}_a^{RUE}-{R}^{\mathrm{t}}{t}_a^{{\prime\prime\prime} }+{\tau}_a^{RUE}\right){v}_a^{{\prime\prime\prime} }-\sum \limits_{a\in A}{\tau}_a^{RUE}{v}_a^{RUE}.\end{array}} $$

(25)

According to Lemma 1 in Appendix 1, the first term on the right side of inequality (25) is smaller than or equal to $\sum \limits_{a\in A}{\tau}_a^{RUE}{v}_a^{RUE}$. Thus, the right side of (25) is smaller than or equal to zero. Because the term in the square bracket in (24) is smaller than or equal to zero, the left side of (24) is smaller than or equal to the second term on the right side of (24):

$$ \sum \limits_{p\in P}{b}_p^{RUE}{f}_p^{RUE}-\sum \limits_{p\in P}{b}_p^{{\prime\prime\prime} }{f}_p^{{\prime\prime\prime}}\le 0+{R}^{\mathrm{u}}\sum \limits_{p\in P}\left({\varsigma}_p^{RUE}-{\varsigma}_p^{{\prime\prime\prime}}\right){f}_p^{{\prime\prime\prime} }. $$

(26)

It is assumed that the mapping ς = (ς_p)_{∀p ∈ P} is monotone in terms of path flow f. Thus, the following holds:

$$ {R}^{\mathrm{u}}\sum \limits_{p\in P}\left({\varsigma}_p^{{\prime\prime\prime} }-{\varsigma}_p^{RUE}\right)\left({f}_p^{RUE}-{f}_p^{{\prime\prime\prime}}\right)\le 0, $$

or equivalently,

$$ {R}^{\mathrm{u}}\sum \limits_{p\in P}{\varsigma}_p^{\prime \prime \prime }{f}_p^{RUE}+{R}^{\mathrm{u}}\sum \limits_{p\in P}\left({\varsigma}_p^{RUE}-{\varsigma}_p^{\prime \prime \prime}\right){f}_p^{\prime \prime \prime}\le {R}^{\mathrm{u}}\sum \limits_{p\in P}{\varsigma}_p^{RUE}{f}_p^{RUE}. $$

Eliminating the non-negative term ${R}^{\mathrm{u}}\sum \limits_{p\in P}{\varsigma}_p^{{\prime\prime\prime} }{f}_p^{RUE}$ from the above inequality, the inequality still holds, i.e.,

$$ {R}^{\mathrm{u}}\sum \limits_{p\in P}\left({\varsigma}_p^{RUE}-{\varsigma}_p^{{\prime\prime\prime}}\right){f}_p^{{\prime\prime\prime}}\le {R}^{\mathrm{u}}\sum \limits_{p\in P}{\varsigma}_p^{RUE}{f}_p^{RUE}. $$

(27)

Based on inequalities (26) and (27), the following is true:

$$ \sum \limits_{p\in P}{b}_p^{RUE}{f}_p^{RUE}-\sum \limits_{p\in P}{b^{{\prime\prime\prime} }{f}^{{\prime\prime\prime}}}_p\le {R}^{\mathrm{u}}\sum \limits_{p\in P}{\varsigma}_p^{RUE}{f}_p^{RUE}. $$

(28)

Based on (23), the following inequality holds:

$$ \sum \limits_{p\in P}{f}_p{b}_p\le \left({R}^{\mathrm{t}}+{\varepsilon}_{\mathrm{max}}{R}^{\mathrm{u}}\right)\cdot \frac{1}{R^{\mathrm{t}}}\cdot \left(\sum \limits_{p\in P}{f}_p{b}_p-{R}^{\mathrm{u}}\sum \limits_{p\in P}{f}_p\cdot {\varsigma}_p\right), $$

or equivalently,

$$ {R}^{\mathrm{u}}\sum \limits_{p\in P}{f}_p\cdot {\varsigma}_p\le \left(1-\frac{R^{\mathrm{t}}}{R^{\mathrm{t}}+{\varepsilon}_{\mathrm{max}}{R}^{\mathrm{u}}}\right)\sum \limits_{p\in P}{f}_p{b}_p. $$

Based on the above, the following inequality holds:

$$ {R}^{\mathrm{u}}\sum \limits_{p\in P}{\varsigma}_p^{RUE}{f}_p^{RUE}\le \left({R}^{\mathrm{u}}{\varepsilon}_{\mathrm{max}}/\left({R}^{\mathrm{t}}+{R}^{\mathrm{u}}{\varepsilon}_{\mathrm{max}}\right)\right)\sum \limits_{p\in P}{b}_p^{RUE}{f}_p^{RUE}. $$

The right side of the above inequality is an upper bound of the right side of (28) and thus is an upper bound of the left side of (28), which gives

$$ \sum \limits_{p\in P}{b}_p^{RUE}{f}_p^{RUE}-\sum \limits_{p\in P}{b}_p^{{\prime\prime\prime} }{f}_p^{{\prime\prime\prime}}\le \left({R}^{\mathrm{u}}{\varepsilon}_{\mathrm{max}}/\left({R}^{\mathrm{t}}+{R}^{\mathrm{u}}{\varepsilon}_{\mathrm{max}}\right)\right)\sum \limits_{p\in P}{b}_p^{RUE}{f}_p^{RUE}, $$

which further gives

$$ \sum \limits_{p\in P}{b}_p^{RUE}{f}_p^{RUE}/\sum \limits_{p\in P}{b}_p^{{\prime\prime\prime} }{f}_p^{{\prime\prime\prime}}\le 1/\left(1-\left({R}^{\mathrm{u}}{\varepsilon}_{\mathrm{max}}/\left({R}^{\mathrm{t}}+{R}^{\mathrm{u}}{\varepsilon}_{\mathrm{max}}\right)\right)\right)=1+\left({R}^{\mathrm{u}}{\varepsilon}_{\mathrm{max}}/{R}^{\mathrm{t}}\right). $$

This completes the proof. ■.

Appendix 4: Proof of Property 2

Proof

Assume ${\mathbf{v}}^{{\prime\prime}}\left({\mathbf{f}}^{{\prime\prime}}\right)={\left({v}_a^{{\prime\prime}}\right)}_{a\in A}$ is the link flow pattern of the path flow pattern ${\mathbf{f}}^{{\prime\prime} }={\left({f}_p^{{\prime\prime}}\right)}_{p\in P}$ that minimizes the TSTCB given y^RUE. Let ${b}_p^{{\prime\prime} }={b}_p\left({\mathbf{v}}^{{\prime\prime}}\left({\mathbf{f}}^{{\prime\prime}}\right),{\mathbf{y}}^{RUE}\right)$.

By definition, the sum of individual path travel cost budgets is larger than or equal to the monetary value of mean TSTT, i.e.,

$$ {R}^{\mathrm{t}}\sum \limits_{a\in A}{t}_a^{RUE}{v}_a^{RUE}\le \sum \limits_{p\in P}{b}_p^{RUE}{f}_p^{RUE}. $$

Multiplying both sides of the above inequality by (1 + ε_maxR^s/R^t), the inequality still holds. That is,

$$ \left({R}^{\mathrm{t}}+{\varepsilon}_{\mathrm{max}}{R}^{\mathrm{s}}\right)\sum \limits_{a\in A}{t}_a^{RUE}{v}_a^{RUE}\le \left(1+{\varepsilon}_{\mathrm{max}}{R}^{\mathrm{s}}/{R}^{\mathrm{t}}\right)\sum \limits_{p\in P}{b}_p^{RUE}{f}_p^{RUE}. $$

The left side of above inequality is an upper bound of the TSTCB according to (22) in Appendix 2. Thus, the right side of the above inequality is larger than or equal to ${TSTCB}_{R^{\mathrm{t}},{R}^{\mathrm{s}}}\left({\mathbf{v}}^{RUE}\left({\mathbf{f}}^{RUE}\right),{\mathbf{y}}^{RUE}\right)$, i.e.,

$$ {TSTCB}_{R^{\mathrm{t}},{R}^{\mathrm{s}}}\left({\mathbf{v}}^{RUE}\left({\mathbf{f}}^{RUE}\right),{\mathbf{y}}^{RUE}\right)\le \left(1+{\varepsilon}_{\mathrm{max}}{R}^{\mathrm{s}}/{R}^{\mathrm{t}}\right)\sum \limits_{p\in P}{b}_p^{RUE}{f}_p^{RUE}. $$

(29)

Similarly, the following inequality holds:

$$ \left({R}^{\mathrm{t}}+{\varepsilon}_{\mathrm{max}}{R}^{\mathrm{u}}\right)\sum \limits_{a\in A}{t}_a^{{\prime\prime} }{v}_a^{{\prime\prime}}\le \left(1+{\varepsilon}_{\mathrm{max}}{R}^{\mathrm{u}}/{R}^{\mathrm{t}}\right){TSTCB}_{R^{\mathrm{t}},{R}^{\mathrm{s}}}\left({\mathbf{v}}^{{\prime\prime}}\left({\mathbf{f}}^{{\prime\prime}}\right),{\mathbf{y}}^{RUE}\right). $$

The left side of the above inequality is an upper bound of the sum of individual path travel cost budgets according to (23) in Appendix 2. Thus, the right side of the above inequality is larger than or equal to $\sum \limits_{p\in P}{b}_p^{{\prime\prime} }{f}_p^{{\prime\prime} }$, which further gives

$$ {TSTCB}_{R^{\mathrm{t}},{R}^{\mathrm{s}}}\left({\mathbf{v}}^{{\prime\prime}}\left({\mathbf{f}}^{{\prime\prime}}\right),{\mathbf{y}}^{RUE}\right)\ge \left(\sum \limits_{p\in P}{b}_p^{{\prime\prime} }{f}_p^{{\prime\prime}}\right)/\left(1+{\varepsilon}_{\mathrm{max}}{R}^{\mathrm{u}}/{R}^{\mathrm{t}}\right). $$

(30)

Dividing the left side of (29) by the left side of (30), and dividing the right side of (29) by the right side of (30), we obtain the following inequality:

$$ \frac{TSTCB_{R^{\mathrm{t}},{R}^{\mathrm{s}}}\left({\mathbf{v}}^{RUE}\left({\mathbf{f}}^{RUE}\right),{\mathbf{y}}^{RUE}\right)}{TSTCB_{R^{\mathrm{t}},{R}^{\mathrm{s}}}\left({\mathbf{v}}^{{\prime\prime}}\left({\mathbf{f}}^{{\prime\prime}}\right),{\mathbf{y}}^{RUE}\right)}\le \frac{\sum \limits_{p\in P}{b}_p^{RUE}{f}_p^{RUE}}{\sum \limits_{p\in P}{b}_p^{{\prime\prime} }{f}_p^{{\prime\prime} }}\left(1+{\varepsilon}_{\mathrm{max}}{R}^{\mathrm{s}}/{R}^{\mathrm{t}}\right)\left(1+{\varepsilon}_{\mathrm{max}}{R}^{\mathrm{u}}/{R}^{\mathrm{t}}\right). $$

(31)

In (31), $\sum \limits_{p\in P}{b}_p^{{\prime\prime} }{f}_p^{{\prime\prime} }$ is larger than $\sum \limits_{p\in P}{b}_p^{{\prime\prime\prime} }{f}_p^{{\prime\prime\prime} }$ defined in Property 1, because $\sum \limits_{p\in P}{b}_p^{{\prime\prime\prime} }{f}_p^{{\prime\prime\prime} }$ is the minimum sum of individual path travel cost budgets given y^RUE. Thus,

$$ \sum \limits_{p\in P}{b}_p^{RUE}{f}_p^{RUE}/\sum \limits_{p\in P}{b}_p^{{\prime\prime} }{f}_p^{{\prime\prime}}\le \sum \limits_{p\in P}{b}_p^{RUE}{f}_p^{RUE}/\sum \limits_{p\in P}{b}_p^{{\prime\prime\prime} }{f}_p^{{\prime\prime\prime} }. $$

Together with Property 1, we obtain the following inequality:

$$ \sum \limits_{p\in P}{b}_p^{RUE}{f}_p^{RUE}/\sum \limits_{p\in P}{b}_p^{{\prime\prime} }{f}_p^{{\prime\prime}}\le 1+{\varepsilon}_{\mathrm{max}}{R}^{\mathrm{u}}/{R}^{\mathrm{t}}. $$

(32)

Inequalities (31) and (32) indicate that the left side of (31) is smaller than or equal to (1 + ε_maxR^s/R^t)(1 + ε_maxR^u/R^t)². This completes the proof.